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2015-09-16T03:55:47+05:30

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Vector A = i + 3 j            magnitude = √(1+9) = √10
vector B = i + j                magnitude = √2

angle between them =  Ф
     Cos Ф= A . B / (√10 * √2)  = (1*1+3*1) / (√20) = 2/√5

Component of vector A along B is    A cosФ = (i + 3 j) * 2/√5
 component of vector B along  A  is    B cos Ф =  (i + j) 2/√5
2 5 2
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